Consider a compact Riemann surface $C$ embedded in a projective plane with coordinates $[x:y:z]$.
Is it possible to write any meromorphic function $f\in M(C)$ as a fraction $f=\frac{P(x,y,z)}{Q(x,y,z)}$ with $P,Q$ homogeneous polynomials of the same degree having no common zero on $C$?
Of course $Q(x,y,z)$ will have zeroes on $C$ since a (non constant) meromorphic function on $C$ can not be holomorphic everywhere, but I'm asking whether at such zeroes $P$ will be non-zero.
I am pretty sure that Riemann-Roch guarantees that the answer is yes provided the common degree of $P$ and $Q$ is large enough, but I'd be grateful for a clean write-up.
Meromorphic functions on a Riemann surface as quotient of polynomials
algebraic-geometryriemann-surfaces
Best Answer
Often no, by Bezout's theorem.
Let $$E:zy^2=x^3+z^3\subset \Bbb{P^2(C)}$$
For an homogeneous polynomial $f\in \Bbb{C}[x,y,z]$ of degree $d$ its zeros on $E$ are well defined, to define their multiplicities: take $w\in \{x,y,z\}$ not vanishing at $P$ and look at the order at $P$ of the zero of the meromorphic function $f/w^d$.
$x$ has 3 simple zeros at $[0:1:1],[0:-1:1],[0:1:0]$
$x^d$ has $3d$ zeros
$f/x^d$ is a meromorphic function, with the same number of zeros and poles, so $f$ must have $3d$ zeros.
The meromorphic function $x/z$ has two simple zeros at $[0:1:1],[0:-1:1]$ and a double pole at $[0:1:0]$.
It can't be that $$x/z=f/g$$ with $f,g$ homogeneous of same degree $d$ and with no common zero, as $f/g$ would have $3d$ zeros and poles whereas $x/z$ has $2$ zeros and poles.